Problem 573 - Advanced Calculus

Question

Suppose that the electrical potential at the point \((\mathrm{x}, \mathrm{y}, \mathrm{z})\) is $$ E(x, y, z)=x^{2}+y^{2}-2 z^{2} $$ What is the direction of the acceleration at the point \((1,3,2) ?\)

Step-by-Step Solution

Answer

The direction of the acceleration at the point \((1,3,2)\) is \(\left(\frac{-2}{\sqrt{104}}, \frac{-6}{\sqrt{104}}, \frac{8}{\sqrt{104}}\right)\).

Step 1: Find the gradient of the potential function

To find the gradient of the potential function, we need to compute partial derivatives of \(E(x, y, z)\) with respect to \(x\), \(y\), and \(z\). The gradient is represented as \(\nabla E(x,y,z) = \left(\frac{\partial E}{\partial x}, \frac{\partial E}{\partial y}, \frac{\partial E}{\partial z}\right)\). $$ \frac{\partial E}{\partial x} = 2x, \quad \frac{\partial E}{\partial y} = 2y, \quad \frac{\partial E}{\partial z} = -4z $$

Step 2: Evaluate the gradient at the given point

Next, we need to evaluate the gradient at the given point \((1, 3, 2)\). We plug the coordinates into the partial derivatives as follows: $$ \frac{\partial E}{\partial x}(1,3,2) = 2(1) = 2, \\ \frac{\partial E}{\partial y}(1,3,2) = 2(3) = 6, \\ \frac{\partial E}{\partial z}(1,3,2) = -4(2) = -8 $$

Step 3: Find the negative gradient

We need to compute the negative gradient, since this will indicate the direction of the acceleration. The negative gradient will be as follows: $$ -\nabla E(1,3,2) = \left(-\frac{\partial E}{\partial x}(1,3,2), -\frac{\partial E}{\partial y}(1,3,2), -\frac{\partial E}{\partial z}(1,3,2) \right) = (-2, -6, 8) $$

Step 4: Normalize the negative gradient

To find the direction of the acceleration, we need to normalize the negative gradient. This involves dividing each component by the magnitude of the vector. The magnitude of the vector is given by \(\lVert \vec{v} \rVert = \sqrt{-2^2 + (-6)^2 + 8^2} = \sqrt{104}\). Now, we can normalize the vector: $$ \frac{(-2, -6, 8)}{\sqrt{104}} = \left(\frac{-2}{\sqrt{104}}, \frac{-6}{\sqrt{104}}, \frac{8}{\sqrt{104}}\right) $$ The direction of the acceleration at the point \((1,3,2)\) is \(\left(\frac{-2}{\sqrt{104}}, \frac{-6}{\sqrt{104}}, \frac{8}{\sqrt{104}}\right)\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, and they help us understand how a function changes with respect to one variable while keeping the others constant. If you imagine a three-dimensional landscape determined by an equation, partial derivatives reveal how steep the slope is in different directions.
For a function like the electrical potential \(E(x, y, z) = x^2 + y^2 - 2z^2\), we have three variables: \(x\), \(y\), and \(z\). To find the gradient, which tells us the direction of greatest increase, we take the partial derivatives with respect to each variable:
\[\frac{\partial E}{\partial x} = 2x, \quad \frac{\partial E}{\partial y} = 2y, \quad \frac{\partial E}{\partial z} = -4z\]
This gives us a vector known as the gradient \(abla E(x, y, z)\), signifying the rate and direction of the change at any point \((x, y, z)\).
Applying this to the specific point \((1, 3, 2)\) gives us the partial derivative values and thus the components of the gradient vector at that point.

Normalization of Vectors

Normalization is the process of converting a vector into a unit vector, meaning it maintains its direction but reduces its length to 1. This is often done when you want to represent direction only, without magnitude.
To normalize a vector \(\vec{v} = (a, b, c)\), calculate its magnitude \(\lVert \vec{v} \rVert\) using the formula:

• \[\lVert \vec{v} \rVert = \sqrt{a^2 + b^2 + c^2}\]

Then, divide each component of the vector by this magnitude:

• \[\left(\frac{a}{\lVert \vec{v} \rVert}, \frac{b}{\lVert \vec{v} \rVert}, \frac{c}{\lVert \vec{v} \rVert}\right)\]

In our example, with the negative gradient vector \((-2, -6, 8)\), its magnitude is \(\lVert \vec{v} \rVert = \sqrt{104}\). Normalize by dividing each component by \(\sqrt{104}\):
\[\left(\frac{-2}{\sqrt{104}}, \frac{-6}{\sqrt{104}}, \frac{8}{\sqrt{104}}\right)\]
This final result represents just the direction of the vector with a magnitude of 1 while keeping its original orientation.

Negative Gradient Direction

The negative gradient direction is particularly interesting when we're discussing optimization problems or scenarios involving potential fields like this one.
The gradient vector \(abla E(x, y, z)\) points in the direction of the steepest increase of the function. Conversely, the negative of this gradient \(-abla E(x, y, z)\) points towards the direction of the steepest decrease. This is why, for minimizing or descending down the slope quickly, we frequently use the negative gradient direction.
Computing the negative gradient involves simply taking the opposite sign of each component in the gradient vector. In our example, the gradient evaluated at \((1, 3, 2)\) yields \((2, 6, -8)\). Thus, the negative gradient becomes \((-2, -6, 8)\).
This indicates that if we're considering the acceleration due to an electrical potential field, moving in the direction of the negative gradient will effectively find the quickest path of descent, akin to a ball rolling downhill, driven by gravity in a parabolic landscape.